Problem: If two of the roots of \[2x^3 + 8x^2 - 120x + k = 0\]are equal, find the value of $k,$ given that $k$ is positive.
Solution: Let the three roots of the equation be $a,$ $a,$ and $b.$ Then by Vieta's formulas, \[\begin{aligned}a+a+b&=-\tfrac82=-4, \\ ab+ab+a^2 &= \tfrac{120}2 = -60. \end{aligned}\]These equations simplify to $2a+b=-4$ and $2ab+a^2=-60.$ From the first equation, we get $b=-4-2a,$ and substituting into the second equation gives \[2a(-4-2a)+a^2=-60,\]or \[3a^2+8a-60=0.\]This factors as \[(a+6)(3a-10)=0,\]so either $a=-6$ or $a=\tfrac{10}{3}.$ If $a=-6$, then $b=-4-2a=8,$ so by Vieta, $k = -2a^2b=-576,$ which is not positive. If $a=\tfrac{10}{3},$ then $b=-4-2a=-\tfrac{32}{3},$ so by Vieta, $k=-2a^2b=\boxed{\tfrac{6400}{27}},$ which is the answer.